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shadowwoelf · Sep 12, 2009, 05:36 PM

I am starting calculus and I just don't understand the point of limits. Why do I care about the values near a point?

example: 1/x. Why should I care about values epsilon of 0?

Also: 1/x when x is 0 does that mean the answer is infinity, correct?

galactus · Sep 13, 2009, 04:49 AM

We care about limits because that is one of the main points in calculus. For instance, when you get to differentiation you find the slope at a point. That is where the limit comes in.

When finding max and min values, we have to find the point where the slope is 0.

Apparently, you are being tortured with the rigorous definition of a limit. I know, it is the most confusing aspect of Calc I. It is not essential to know this formal definition to continue in calculus. If one ever takes analysis, then it is important.

Let me try to explain this rigorous definition in a simpler way. In math, many times a rather simple concept can be made confusing with all the wacky notation.

Here is the definition you seen:

$\lim_{x\to a}f(x)=L$

if and only if, given ${\epsilon}>0, \exists {\delta}>0$ such that if $0<|x-a|<{\delta}$ , then $|f(x)-L|<{\epsilon}$

Is that it? Kind of confusing for a new calc student, huh?

Pictures would help here, so bear with me. I may have to come back later.

Translation 1: Given ${\epsilon}$ , a small positive number, we can always find ${\delta}$ , another small positive number, such that if x is within a distance ${\delta}$ from but not exactly at 'a', then f(x) is a within a distance ${\epsilon}$ from L.

Translation 2:

Interpret |x-a| as the distance between x and a, but instead of the one-dimensional picture it really is, imagine that there is a circle around a point 'a' of radius $\delta$ . $|x-a|<{\delta}$ stands for all points inside this circle.
Similarly, imagine a circle of radius $\epsilon$ around L, with $|f(x)-L|<{\epsilon}$ the set of all points f(x) that are inside this circle.

This definition says given $\epsilon>0$ (given a circle of radius $\epsilon$ around L), we can find ${\delta}>0$ (circle of radius $\delta$ around a) such that if $0<|x-a|<{\delta}$ (if we take any x inside the circle, then $|f(x)-L|<{\epsilon}$ (f(x)) will be inside the circle of radius $\epsilon$ but not exactly at L.

Now take another $\epsilon$ , call it ${\epsilon}_{2}$ , positive but smaller than $\epsilon$ (a smaller circle around L); there exists another $\delta, \;\ {\delta}_{2}$ , usually a smaller circle around a, such that if $0<|x-a|,{\delta}_{2}$ , then $|f(x)-L|<{\epsilon}_{2}$

Take them smaller and smaller. The smaller the epsilon, the smaller the delta. f(x) goes to L as x goes to a.

It is an approximating process that many times gives exact answers.

For your problem $\lim_{x\to 0}\frac{1}{x}$ . The limit does not exist. That is because if we approach 0 from the left we get $-\infty$ and of we approach from the right we get $+\infty$ .

Try plugging in smaller and smaller numbers into 1/x and see. Try x=.1, 01, 001, 0001, and so on. Then, try -.1, -.01, -.001, -.0001, and so on. As the numbers get smaller, 1/x gets larger and larger. But rather its negative or positive depends on what direction you approach from.

I am sorry, this is long winded. But I hope it helps.

Here are 3 diagrams I hope help some.